- #1

sponsoredwalk

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My book is talking about linear independence.

As I understand the concept it means that a vector has to be in a different direction, i.e. non-collinear, with the vector in comparison.

Mathematically, my book has defined linearly independent vectors as;

[tex] a \overline{u} \ + \ b \overline{v} \ = \ 0 \ with \ a \ = \ 0 \ and \ b = \ 0 [/tex]

The book says that linearly dependent vectors are those in which the scalars a & b in the above are not equal to zero.

The reason is that you could solve for either vector, if either scalar was non-zero, and then describe one vector as a scalar multiple of the other.

Okay, that seems about right.

My problem comes from the following;

[tex] assume \ \overline{u} \ and \ \overline{v} \ to \ be \ linearly \ independent [/tex]

[tex] a \overline{u} \ + \ b \overline{v} \ = \ \alpha \overline{u} \ + \ \beta \overline{v} [/tex]

[tex] ( \ a \ - \ \alpha \ ) \overline{u} \ + \ ( \ b \ - \ \beta \ ) \overline{v} \ = \ 0 [/tex]

This must imply that a=α & that b=β.

This must also imply that a-α=0 & that b-β=0.

But shouldn't a & α, b & β, all have to be equal to zero anyway?

Isn't the above saying that zero minus zero equals zero?

**If all the scalar coefficients are not equal to zero then the vectors must be linearly dependent, I get the feeling that we are trying to define a way to say that you can set a vector to have non-zero coefficents if you set it equal to itself & reference it to itself. But if you do this aren't you defining linearly dependent vectors simply by the fact that the scalars are non-zero?**

What am I not realizing?What am I not realizing?

Please go easy, I don't know anything about basis or anything as the book is trying to define the concept using the above principle.